Problem: Nauuo is a girl who loves playing chess.

One day she invented a game by herself which needs n chess pieces to play on a m× m chessboard. The rows and columns are numbered from 1 to m. We denote a cell on the intersection of the r-th row and c-th column as (r,c).

The game's goal is to place n chess pieces numbered from 1 to n on the chessboard, the i-th piece lies on (r_i,\,c_i), while the following rule is satisfied: for all pairs of pieces i and j, |r_i-r_j|+|c_i-c_j|≥|i-j|. Here |x| means the absolute value of x.

However, Nauuo discovered that sometimes she couldn't find a solution because the chessboard was too small.

She wants to find the smallest chessboard on which she can put n pieces according to the rules.

She also wonders how to place the pieces on such a chessboard. Can you help her?

Input

The only line contains a single integer n (1≤ n≤ 1000) — the number of chess pieces for the game.

Output

The first line contains a single integer — the minimum value of m, where m is the length of sides of the suitable chessboard.

The i-th of the next n lines contains two integers r_i and c_i (1≤ r_i,c_i≤ m) — the coordinates of the i-th chess piece.

If there are multiple answers, print any.

Examples

Input


2


Output


2
1 1
1 2

Input


4


Output


3
1 1
1 3
3 1
3 3

Note

In the first example, you can't place the two pieces on a 1×1 chessboard without breaking the rule. But you can place two pieces on a 2×2 chessboard like this:

<image>

In the second example, you can't place four pieces on a 2×2 chessboard without breaking the rule. For example, if you place the pieces like this:

<image>

then |r_1-r_3|+|c_1-c_3|=|1-2|+|1-1|=1, |1-3|=2, 1<2; and |r_1-r_4|+|c_1-c_4|=|1-2|+|1-2|=2, |1-4|=3, 2<3. It doesn't satisfy the rule.

However, on a 3×3 chessboard, you can place four pieces like this:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool isprime(int64_t n) {
  for (int64_t i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
int64_t factorial(int64_t n) {
  return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
}
int64_t gcd(int64_t a, int64_t b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int64_t lcm(int64_t a, int64_t b) { return (a / gcd(a, b) * b); }
int64_t max(int64_t a, int64_t b) {
  int64_t sol = a > b ? a : b;
  return sol;
}
int64_t min(int64_t a, int64_t b) {
  int64_t sol = a < b ? a : b;
  return sol;
}
int64_t power(int64_t x, int64_t y) {
  int64_t ans = 1;
  while (y > 0) {
    if (y & 1) {
      ans *= x;
    }
    y >>= 1LL;
    x *= x;
  }
  return ans;
}
int64_t mod = 1000000007;
inline int64_t solve() {
  int64_t n;
  cin >> n;
  if (n == 1) {
    cout << 1 << "\n";
    cout << 1 << " " << 1 << "\n";
    return 0;
  }
  int64_t x = n / 2 + 1;
  cout << x << "\n";
  int64_t cnt = 0;
  for (int64_t i = 1; i <= x; ++i) {
    cout << 1 << " " << i << "\n";
    cnt++;
    if (cnt == n / 2) {
      break;
    }
  }
  for (int64_t i = 1; i <= x; ++i) {
    cout << x << " " << i << "\n";
    cnt++;
    if (cnt == n) {
      break;
    }
  }
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  solve();
}